Soluciones RD Sharma Clase 11 – Capítulo 29 Límites – Ejercicio 29.7 | conjunto 2

Pregunta 32. lim x→0 [{sin(a + x) + sin(a – x) – 2sina}/(xsenx)]

Solución:

Tenemos,

lím x→0 [{sin(a + x) + sin(a – x) – 2sina}/(xsenx)]

\lim_{x\to0}[\frac{2sin(\frac{a+x+a-x}{2})sin(\frac{a+x-a+x}{2})}{xsinx}]

=\lim_{x\to0}[\frac{2sina(cosx-1)}{xsinx}]

=\lim_{x\to0}[\frac{-2sina(1-cosx)}{xsinx}]

-2sina\lim_{x\to0}[\frac{2sin^2\frac{x}{2}}{x.2sin\frac{x}{2}cos\frac{x}{2}}]

-2sina\lim_{x\to0}[\frac{sin\frac{x}{2}}{x.cos\frac{x}{2}}]

-2sina\lim_{x\to0}[\frac{tan\frac{x}{2}}{x}]

-2sina\lim_{x\to0}[\frac{tan\frac{x}{2}}{\frac{x}{2}×2}]

= -2sina × (1/2)

= -sina

Pregunta 33. lím x→0 [{x 2 – tan2x}/(tanx)]

Solución:

Tenemos,

lím x→0 [{x 2 -tan2x}/(tanx)]

Dividiendo numerador por 2x y denominador por x.

=\lim_{x\to0}[\frac{(\frac{x^2}{2x}-\frac{tan2x}{2x})×2x}{\frac{tanx}{x}×x}]

=\lim_{x\to0}[\frac{2(\frac{x}{2}-\frac{tan2x}{2x})}{\frac{tanx}{x}}]

= 2(0 – 1)/1

= -2

Pregunta 34. lím x→0 [{√2 – √(1 + cosx)}/x 2 ]

Solución:

Tenemos,

límite x→0 [{√2 – √(1 + cosx)}/x 2 ]

Sobre la racionalización del numerador

= lím x→0 [{2-(1+cosx)}/x 2 {√2+√(1+cosx)}]

= lím x→0 [(1-cosx))/x 2 {√2+√(1+cosx)}]

\lim_{x\to0}[\frac{sin^2(\frac{x}{2})}{x^2.(\sqrt{2}+\sqrt{1+cosx}}]

\lim_{x\to0}[\frac{2sin^2(\frac{x}{2})}{(\frac{x}{2})^2×4}×\frac{1}{(\sqrt{2}+\sqrt{1+cosx}}]

= 2 × (1/4) × [1/{√2 + √(1 + 1)}]

= (2/4) × (1/2√2)

= (1/4√2)

Pregunta 35. lim x→0 [{xtanx}/(1 – cosx)]

Solución:

Tenemos,

límite x→0 [{xtanx}/(1 – cosx)]

Al dividir el numerador y el denominador por x 2

=\lim_{x\to0}[\frac{\frac{tanx}{x}}{\frac{2sin^2(\frac{x}{2})}{x^2}}]

=\lim_{x\to0}[\frac{\frac{tanx}{x}}{\frac{2sin^2(\frac{x}{2})}{(\frac{x}{2})^2×4}}]

x→0 x→0

= (4/2)

= 2

Pregunta 36. lim x→0 [{x 2 + 1 – cosx}/(xsenx)]

Solución:

Tenemos,

límite x→0 [{x 2 + 1 – cosx}/(xsenx)]

= lím x→0 [{x 2 + 2sen 2 (x/2)}/(xsenx)]

Al dividir el numerador y el denominador por x 2

\lim_{x\to0}[\frac{1+\frac{2sin^2\frac{x}{2}}{x^2}}{{\frac{sinx}{x}}}]

\lim_{x\to0}[\frac{1+2(\frac{sin\frac{x}{2}}{\frac{x}{2}×2})^2}{{\frac{sinx}{x}}}]

x→0

\frac{1+\frac{2}{4}}{1}

= 3/2

Pregunta 37. lím x→0 [sen2x{cos3x – cosx}/(x 3 )]

Solución:

Tenemos,

lím x→0 [sen2x{cos3x – cosx}/(x 3 )]

\lim_{x\to0}\frac{(sin2x)×[-2sin(\frac{3x+x}{2})sin(\frac{3x-x}{2})]}{x^3}

-2\lim_{x\to0}(\frac{sin2x}{2x})×(\frac{sin2x}{2x})×(\frac{sinx}{x})×2×2

x→0

= -2 × 2 × 2

= -8

Pregunta 38. lím x→0 [{2senx° – sen2x°}/(x 3 )]

Solución:

Tenemos,

lím x→0 [{2senx°-sen2x°}/(x 3 )]

= lím x→0 [{2senx°-2senx°cosx°}/(x 3 )]

= lím x→0 [2senx°{1-cosx°}/(x 3 )]

= lím x→0 [2senx°{2sen 2 (x°/2)}/(x 3 )]

4\lim_{x\to0}\frac{sin\frac{πx}{180}×sin^2(\frac{πx}{360})}{x.x^2}

4\lim_{x\to0}\frac{sin\frac{πx}{180}}{\frac{πx}{180}}×(\frac{sin(\frac{πx}{360})}{\frac{πx}{360}})^2×\frac{π}{180}×(\frac{π}{360})^2

= 4 × [π 3 /(180 × 360 × 360)]

= (π/180) 3

Pregunta 39. lim x→0 [{x 3 .cotx}/(1 – cosx)]

Solución:

Tenemos,

límite x→0 [{x 3 .cotx}/(1 – cosx)]

= lím x→0 [x 3 /{tanx(1 – cosx)}]

\lim_{x\to0}[\frac{x}{tanx}×\frac{x^2}{2sin^2\frac{x}{2}}]

\lim_{x\to0}[\frac{x}{tanx}×\frac{\frac{x^2}{4}×4}{2sin^2\frac{x}{2}}]

\lim_{x\to0}[\frac{x}{tanx}×(\frac{\frac{x}{2}}{sin\frac{x}{2}})^2×2]

x→0 x→0

= 2

Pregunta 40. lim x→0 [{x.tanx}/(1 – cos2x)]

Solución:

Tenemos,

lím x→0 [{x.tanx}/(1 – cos2x)]

= lim x→0 [{x.tanx}/(2sen 2 x)]

Al dividir el numerador y el denominador por x 2

\lim_{x\to0}[\frac{\frac{tanx}{x}}{2\frac{sin^2x}{x^2}}]

x→0 x→0

= (1/2)

Pregunta 41. lim x→0 [{sin(3 + x) – sin(3 – x)}/x]

Solución:

Tenemos,

lím x→0 [{sen(3 + x) – sin(3 – x)}/x]

=\lim_{x\to0}[\frac{2cos(\frac{3+x+3-x}{2})sin(\frac{3+x-3+x}{2})}{x}]

= 2Lím x→0 [cos3.senx/x]

= 2cos × 3lim x→0 [senx/x]

x→0

= 2cos3

Pregunta 42. lím x→0 [{cos2x – 1)}/(cosx – 1)]

Solución:

Tenemos,

límite x→0 [{cos2x – 1)}/(cosx – 1)]

= lím x→0 [(2sen 2 x)/{2sen 2 (x/2)}]

= lím x→0 [(sen 2 x)/{sen 2 (x/2)}]

\lim_{x\to0}[\frac{sin^2x}{x^2}×x^2][\frac{1}{\frac{sin^2\frac{x}{2}}{(\frac{x}{2})^2}×(\frac{x}{2})^2}]

x→0

= (x 2 ) × (4/x 2 )

= 4

Pregunta 43. lím x→0 [{3sen 2 x – 2senx 2 )}/(3x 2 )]

Solución:

Tenemos,

límite x→0 [{3sen 2 x – 2senx 2 )}/(3x 2 )]

= límite x→0 [(3sen 2 x/3x 2 ) – (2senx 2 /3x 2 )]

x→0

= 1 – 2/3

= (3 – 2)/3

= (1/3)

Pregunta 44. lim x→0 [{√(1 + senx) – √(1 – senx)}/x]

Solución:

Tenemos,

lím x→0 [{√(1 + senx) – √(1 – senx)}/x]

Sobre la racionalización del numerador.

= lim x→0 [{(1 + senx) – (1 – senx)}/x{√(1 + senx) + √(1 – senx)}]

= lim x→0 [2(senx)/x{√(1 + senx) + √(1 – senx)}]

2\lim_{x\to0}[\frac{sinx}{x}×\frac{1}{\sqrt{1+sinx}+\sqrt{1-sinx}}]

x→0

= 2 × {1/(√1 + √1)}

= 2/2

= 1

Pregunta 45. lím x→0 [(1 – cos4x)/x 2 ]

Solución:

Tenemos,

lím x→0 [(1 – cos4x)/x 2 ]

= lím x→0 [2sen 2 2x/x 2 ]

2\lim_{x\to0}[(\frac{sin2x}{2x})^2×4]

x→0

= 2 × 4

= 8

Pregunta 46. lim x→0 [(xcosx + senx)/(x 2 + tanx)]

Solución:

Tenemos,

lím x→0 [(xcosx + senx)/(x 2 + tanx)]

= lím x→0 [x(cosx+senx/x)/x(x + tanx/x)]

= lím x→0 [(cosx + senx/x)/(x + tanx/x)]

\lim_{x\to0}\frac{\lim_{x\to0}cosx+\lim_{x\to0}\frac{sinx}{x}}{\lim_{x\to0}x+\lim_{x\to0}\frac{tanx}{x}}

x→0

= (1 + 1)/(1 + 0) 

= 2

Pregunta 47. lím x→0 [(1 – cos2x)/(3tan 2 x)]

Solución:

Tenemos,

lím x→0 [(1 – cos2x)/(3tan 2 x)]

= limx→0[2sen 2 x/3tan 2 x]

=\frac{2}{3}\lim_{x\to0}\frac{sin^2x}{\frac{sin^2x}{cos^2x}}

= (2/3)lim x→0 [cos 2 x]

= (2/3)

Pregunta 48. lím θ→0 [(1 – cos4θ)/(1 – cos6θ)]

Solución:

Tenemos,

lím θ→0 [(1 – cos4θ)/(1 – cos6θ)]

= lím θ→0 [2sen 2 2θ/2sen 2 3θ]

= lím θ→0 [sen 2 2 θ/sen 2 3 θ]

=\lim_{θ\to0}[\frac{sin^22θ}{(2θ)^2}×(2θ)^2×\frac{1}{\frac{sin^23θ}{(3θ)^2}×(3θ)^2}]

= [(4θ 2 )/(9θ 2 )]

= (4/9)

Pregunta 49. lím x→0 [(ax + xcosx)/(bsinx)]

Solución:

Tenemos,

lím x→0 [(ax + xcosx)/(bsinx)]

Al dividir el numerador y el denominador por x

\lim_{x\to0}\frac{(a+cosx)}{\frac{bsinx}{x}}

x→0

(a + cos 0)/b × 1

= (a + 1)/b

Pregunta 50. lím θ→0 [(sin4θ)/(tan3θ)]

Solución:

Tenemos,

lím θ→0 [(sen4θ)/(tan3θ)]

=\lim_{θ\to0}[\frac{sin4θ}{(4θ)}×(4θ)×\frac{1}{\frac{tan3θ}{(3θ)}×(3θ)}]

x→0 x→0

= (4θ/3θ)

= (4/3)

Pregunta 51. lím x→0 [{2senx – sen2x}/(x 3 )]

Solución:

Tenemos,

límite x→0 [{2senx – sen2x}/(x 3 )]

= lím x→0 [{2senx – 2senxcosx}/(x 3 )]

= lím x→0 [2senx{1 – cosx}/(x 3 )]

= lím x→0 [2senx{2sen 2 (x/2)}/(x 3 )]

4\lim_{x\to0}\frac{sinx}{x}×\frac{sin^2\frac{x}{2}}{(\frac{x}{2})^2×4}

x→0

= (4/4)

= 1

Pregunta 52. lím x→0 [{1 – cos5x}/{1 – cos6x}]

Solución:

Tenemos,

límite x→0 [{1 – cos5x}/{1 – cos6x}]

\lim_{x\to0}[\frac{2sin^2(\frac{5x}{x})}{2sin^23x}]

=\lim_{x\to0}[\frac{sin^2(\frac{5x}{2})}{(\frac{5x}{2})^2}×\frac{25x^2}{4}×\frac{1}{\frac{sin^23x}{(3x)^2}×9x^2}]

x→0

= 25/(4 × 9)

= (25/36)

Pregunta 53. lim x→0 [(cosecx – cotx)/x]

Solución:

Tenemos,

lím x→0 [(cosecx – cotx)/x]

= lím x→0 [(1/senx – cosx/senx)/x]

= lím x→0 [(1 – cosx)/x.senx]

= lím x→0 [2sen 2 (x/2)/x.senx]

2\lim_{x\to0}[\frac{sin^2(\frac{x}{2})}{(\frac{x}{2})^2}×(\frac{x^2}{4})×\frac{1}{\frac{xsinx}{x^2}×x^2}]

x→0

= 2/4

= 1/2

Pregunta 54. lím x→0 [(sin3x + 7x)/(4x + sin2x)]

Solución:

Tenemos,

lím x→0 [(sen3x + 7x)/(4x + sen2x)]

\lim_{x\to0}[\frac{\frac{sin3x}{3x}×3x+7x}{\frac{sin2x}{2x}×2x+4x}]

\lim_{x\to0}[\frac{x(\frac{sin3x}{3x}×3+7)}{x(\frac{sin2x}{2x}×2+4)}]

\lim_{x\to0}[\frac{(\frac{sin3x}{3x}×3+7)}{(\frac{sin2x}{2x}×2+4)}]

x→0

= (7 + 3)/(4 + 2)

= 10/6

= 5/3

Pregunta 55. lím x→0 [(5x + 4sin3x)/(4sin2x + 7x)]

Solución:

Tenemos,

límite x→0 [(5x + 4sin3x)/(4sin2x + 7x)]

=\lim_{x\to0}[\frac{5x+4\frac{sin3x}{3x}×3x}{4\frac{sin2x}{2x}×2x+7x}]

=\lim_{x\to0}[\frac{x(5+4\frac{sin3x}{3x}×3)}{x(4\frac{sin2x}{2x}×2+7)}]

=\lim_{x\to0}[\frac{(5+4\frac{sin3x}{3x}×3)}{(4\frac{sin2x}{2x}×2+7)}]

x→0

= (5 + 4 × 3)/(4 × 2 + 7)

= (17/15)

Pregunta 56. lím x→0 [(3senx – sen3x)/x 3 ]

Solución:

Tenemos,

lím x→0 [(3senx – sen3x)/x 3 ]

= lím x→0 [{3senx – (3senx – 4sen 3 x)/x 3 ]

= lím x→0 [(4sen 3 x)/x 3 ]

= 4Lim x→0 [{(senx)/x} 3 ]

x→0

= 4 × 1

= 4

Pregunta 57. lím x→0 [(tan2x – sin2x)/x 3 ]

Solución:

Tenemos,

lím x→0 [(tan2x – sin2x)/x 3 ]

= lím x→0 [(sen2x/cos2x-sen2x)/x 3 ]

\lim_{x\to0}[\frac{sin2x(\frac{1}{cos2x}-1)}{x^3}]

\lim_{x\to0}[\frac{sin2x(1-cos2x)}{cos2x.x^3}]

= lím x→0 [(2sen2x.sen 2 x)/(x 3 cos2x)]

2\lim_{x\to0}[\frac{sin2x}{2x}×2×(\frac{sinx}{x})^2×\frac{1}{cos2x}]

x→0

= 2 × 2/cos0

= 4

Pregunta 58. lim x→0 [(sinax + bx)/(ax + senbx)]

Solución:

Tenemos,

lím x→0 [(sinax + bx)/(ax + senbx)]

\lim_{x\to0}[\frac{\frac{sinax}{ax}×ax+bx}{ax+\frac{sinbx}{bx}×bx}]

\lim_{x\to0}[\frac{x(\frac{sinax}{ax}×x+b)}{x(a+\frac{sinbx}{bx}×b)}]

\lim_{x\to0}[\frac{(\frac{sinax}{ax}×x+b)}{(a+\frac{sinbx}{bx}×b)}]

x→0

= (1 × a + b)/(a + 1 × b)

= (a + b)/(a + b)

= 1

Pregunta 59. lím x→0 [cosecx-cotx]

Solución:

Tenemos,

lím x→0 [cosecx – cotx]

= lím x→0 [1/senx – cosx/senx]

= lím x→0 [(1 – cosx)/senx]

= lím x→0 [{2sen 2 (x/2)}/{2sen(x/2)cos(x/2)}]

= lím x→0 [sen(x/2)/cos(x/2)]

= límite x→0 [tan(x/2)/ x/2] × x/2

x→0

= 0

Pregunta 60. lim x→0 [{sin(α + β)x + sin(α – β)x + sin2αx}/{cos 2 βx – cos 2 αx}]

Solución:

Tenemos,

 lím x→0 [{sin(α + β)x + sin(α – β)x + sin2αx}/{cos 2 βx – cos 2 αx}]

=\lim_{x\to0}[\frac{2sin(\frac{αx+βx+αx-βx}{2})cos(\frac{αx+βx-αx+βx}{2})+2sinαx.cosαx}{(1-sin^2βx)-(1-sin^2αx)}]

= lim x→0 [{2sinαx.cosβx + 2sinαx.cosαx}/(sen 2 αx – sen 2 βx)]

= lím x→0 [{2sinαx(cosβx + cosαx)}/(sen 2 αx – sen 2 βx)]

2\lim_{x\to0}[\frac{\frac{sinαx}{αx}×αx×(cosβx+cosαx)}{\frac{sin^2αx}{α^2x^2}×α^2x^2-\frac{sin^2βx}{β^2x^2}×β^2x^2}]

\frac{\lim_{x\to0}\frac{sinαx}{αx}\times \lim_{x\to0}(cosβx+cosαx)}{\lim_{x\to0}\frac{sin^2αx}{α^2x^2}×α^2x-\lim_{x\to0}\frac{sin^2βx}{β^2}×β^2x^2} \times \frac{x}{x^2}

x→0

= [{2 × α × 1 × (1 + 1)}/(α2 – β2)] × (1/0)

= (1/0)

= ∞

Pregunta 61. lím x→0 [(cosax – cosbx)/(cosecx – 1)]

Solución:

Tenemos,

lím x→0 [(cosax – cosbx)/(cosecx – 1)]

=\lim_{x\to0}[\frac{-2sin(\frac{ax+bx}{2})sin(\frac{ax-bx}{2})}{-2sin^2(\frac{cx}{2})}]

=\lim_{x\to0}[\frac{sin(\frac{ax+bx}{2})sin(\frac{ax-bx}{2})}{sin^2(\frac{cx}{2})}]

=\lim_{x\to0}[\frac{\frac{sin(\frac{ax+bx}{2})}{(\frac{ax+bx}{2})}×(\frac{ax+bx}{2})×\frac{sin(\frac{ax-bx}{2})}{(\frac{ax-bx}{2})}×\frac{ax-bx}{2}}{(\frac{sin\frac{cx}{2}}{\frac{cx}{2}})^2×(\frac{cx}{2})^2}]

= [(a + b)(a – b)/c 2 ] × (4/4)

= (a 2 – b 2 )/c 2

Pregunta 62. lim h→0 [{(a + h) 2 sin(a + h) – a 2 sina}/h]

Solución:

Tenemos,

lim h→0 [{(a + h) 2 sin(a + h) – a 2 sina}/h]

= lim h→0 [{(a+h) 2 (sina.cosh)+(a+h) 2 (cosa.sinh)-a 2 sina}/h]

= lim h→0 [{(a 2 +2ah+h 2 )(sina.cosh)-a 2 sina+(a+h)2(cosa.sinh)}/h]

= lim h→0 [{a 2 sina(cosh-1)+2ah.sina.cosh+h 2 sina.cosh+(a+h) 2 cosa.sinh}/h]

= lim h→0 [{a 2 sina(-2sin 2 (h/2))+2ah.sina.cosh+h 2 sina.cosh+(a+h) 2 cosa.sinh}/h]

=\lim_{h\to0}[\frac{-a^2sina*sin^2(\frac{h}{2})}{\frac{h}{2}}+2asina.cosh+hsina.cosh+(a+h)^2cosa.sinh]

= 0 + 2asina + 0 + a 2 cosa

= 2a + a 2 cosa

Pregunta 63. Si lim x→0 [kx.cosecx] = lim x→0 [x.coseckx], encuentre K.

Solución:

Tenemos,

límite x→0 [ kx.cosecx ] = límite x→0 [x.cosecx]

límite x→0 [kx/senx] = límite x→0 [x/sinkx]

klim x→0 [x/senx] = lim x→0 [kx/sinkx](1/k)

k = (1/k)

k2 = 1

k = ±1

Publicación traducida automáticamente

Artículo escrito por vivekray59 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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